In each of Examples 8.10, 8.11, and 8.12 (Section 8.4), verify that the relative velocity vector of the two bodies has the same magnitude before and after the collision. In each case, what happens to the ?direction? of the relative velocity vector?

Solution 11DQ Introduction In all cases the velocity before and after collisions are calculated. So we will just calculate the relative velocities and see how the magnitude changes. We will also see how the direction changes before and after collision. Step 1 In example 8.10 the initial velocity of the block A and B are V Ai= 2 m/s and v bi 2 m/s Hence the relative velocity B with respect to A is V BA,i= V Ai+ V Bi = 2.00 m/s + ( 2.00 m/s) = 4.00 m/s And the final velocities are v Af= 1 m/s and v Bf = 3 m/s Hence the relative velocity of B with respect to A in after collision is v BA,f= V Af + V Bf = ( 1 m/s) + (3 m/s) = 4 m/s. Hence the magnitude remain same. Step 2 In the example of 8.11, the initial velocities of neutron and carbon atoms are v = 2.6 × 10 m/s and v = 0 ni ci And the final velocities are 7 7 v nf= 2.2 × 10 m/s and v = ci4 × 10 m/s Hence the initial relative velocity of neutron with respect to carbon is v cni= v civ =ni0 m/s) + (2.6 × 10 m/s) = 2.6 × 10 m/s 7 Now the final relative velocity is 7 7 6 v cnf= v cfv nf = (0.4 × 10 m/s) + ( 2.2 × 10 m/s) = 2.6 × 10 m/s So the magnitude of the relative velocity remains same. Step 4 In the example 8.12, the initial velocity of the mass A is v = 4.00 m/s along the x axis and the initla a1 velocity of mass B is v = 0. hence the initial relative velocity of A with respect to B is b1 V ba = v b1+ v a1 = 0 + 4.00 m/s = 4.00 m/s The final velocity of A is 2.00 m/s at an angle 36.9 degree from x axis. So in the component form it can be written as v = (2.00 m/s)cos(36.9°)i + (2.00 m/s)sin(36.9°)j = (1.60 m/s)i + (1.20 m/s)j ˆ af Now the final velocity of B is 4.47 m/s and at an angle 26.6° below the positive x axis. Hence in component form we can write that v bf= (4.47 m/s)cos( 26.6°)i + (4.47 m/s)sin( 26.6°)j = (4.00 m/s)i (2.00 m/s)j ˆ Hence the relative velocity A with respect to B is ˆ ˆ ˆ ˆ ˆ ˆ v ba,i= v bf+ v af= [(4.00 m/s)i + (2.00 m/s)j] + [(1.60 m/s)i (1.20 m/s)j] = ( 2.4 m/s)i + ( 3.2 m/s)j Hence the magnitude of the velocity is 2 2 |vba,f| = ( 2.4 m/s) + ( 3.2 m/s) = 4.00 m/s Hence the magnitude remains same. Step 4 As we can see from the sign of the relative velocities, the direction in the first and second example become opposite to the initial direction. In the second case the direction is = tan (1 3.2) = 53.1° 2.4 So in the third case the direction is 53.1° above the x axis.